LeetCode #828 — HARD

Count Unique Characters of All Substrings of a Given String

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Let's define a function countUniqueChars(s) that returns the number of unique characters in s.

For example, calling countUniqueChars(s) if s = "LEETCODE" then "L", "T", "C", "O", "D" are the unique characters since they appear only once in s, therefore countUniqueChars(s) = 5.

Given a string s, return the sum of countUniqueChars(t) where t is a substring of s. The test cases are generated such that the answer fits in a 32-bit integer.

Notice that some substrings can be repeated so in this case you have to count the repeated ones too.

Example 1:

Input: s = "ABC"
Output: 10
Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".
Every substring is composed with only unique letters.
Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10

Example 2:

Input: s = "ABA"
Output: 8
Explanation: The same as example 1, except countUniqueChars("ABA") = 1.

Example 3:

Input: s = "LEETCODE"
Output: 92

Constraints:

1 <= s.length <= 10⁵
s consists of uppercase English letters only.

Step 01

Brute Force Baseline

Problem summary: Let's define a function countUniqueChars(s) that returns the number of unique characters in s. For example, calling countUniqueChars(s) if s = "LEETCODE" then "L", "T", "C", "O", "D" are the unique characters since they appear only once in s, therefore countUniqueChars(s) = 5. Given a string s, return the sum of countUniqueChars(t) where t is a substring of s. The test cases are generated such that the answer fits in a 32-bit integer. Notice that some substrings can be repeated so in this case you have to count the repeated ones too.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Dynamic Programming

Example 1

"ABC"

Example 2

"ABA"

Example 3

"LEETCODE"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #828: Count Unique Characters of All Substrings of a Given String
class Solution {
    public int uniqueLetterString(String s) {
        List<Integer>[] d = new List[26];
        Arrays.setAll(d, k -> new ArrayList<>());
        for (int i = 0; i < 26; ++i) {
            d[i].add(-1);
        }
        for (int i = 0; i < s.length(); ++i) {
            d[s.charAt(i) - 'A'].add(i);
        }
        int ans = 0;
        for (var v : d) {
            v.add(s.length());
            for (int i = 1; i < v.size() - 1; ++i) {
                ans += (v.get(i) - v.get(i - 1)) * (v.get(i + 1) - v.get(i));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #828: Count Unique Characters of All Substrings of a Given String
func uniqueLetterString(s string) (ans int) {
	d := make([][]int, 26)
	for i := range d {
		d[i] = []int{-1}
	}
	for i, c := range s {
		d[c-'A'] = append(d[c-'A'], i)
	}
	for _, v := range d {
		v = append(v, len(s))
		for i := 1; i < len(v)-1; i++ {
			ans += (v[i] - v[i-1]) * (v[i+1] - v[i])
		}
	}
	return
}

# Accepted solution for LeetCode #828: Count Unique Characters of All Substrings of a Given String
class Solution:
    def uniqueLetterString(self, s: str) -> int:
        d = defaultdict(list)
        for i, c in enumerate(s):
            d[c].append(i)
        ans = 0
        for v in d.values():
            v = [-1] + v + [len(s)]
            for i in range(1, len(v) - 1):
                ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i])
        return ans

// Accepted solution for LeetCode #828: Count Unique Characters of All Substrings of a Given String
impl Solution {
    pub fn unique_letter_string(s: String) -> i32 {
        let mut d: Vec<Vec<i32>> = vec![vec![-1; 1]; 26];
        for (i, c) in s.chars().enumerate() {
            d[(c as usize) - ('A' as usize)].push(i as i32);
        }
        let mut ans = 0;
        for v in d.iter_mut() {
            v.push(s.len() as i32);
            for i in 1..v.len() - 1 {
                ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
            }
        }
        ans as i32
    }
}

// Accepted solution for LeetCode #828: Count Unique Characters of All Substrings of a Given String
function uniqueLetterString(s: string): number {
    const d: number[][] = Array.from({ length: 26 }, () => [-1]);
    for (let i = 0; i < s.length; ++i) {
        d[s.charCodeAt(i) - 'A'.charCodeAt(0)].push(i);
    }
    let ans = 0;
    for (const v of d) {
        v.push(s.length);

        for (let i = 1; i < v.length - 1; ++i) {
            ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.